The method of solving "by
substitution" works by solving one of the equations (you choose which
one) for one of the variables (you choose which one), and then plugging
this back into the other equation, "substituting" for the chosen
variable and solving for the other. Then you back-solve for the first
variable.

Here is how it works. (I'll
use the same systems as were in a previous page.)

Solve the following
system by substitution.

2x – 3y = –2
4x + y = 24

The idea here is to solve one of the
equations for one of the variables, and plug this into the other equation.
It does not matter which equation or which variable you pick. There
is no right or wrong choice; the answer will be the same, regardless.
But — some choices may be better than others.

For instance, in this case, can you see
that it would probably be simplest to solve the second equation for
"y =", since there is already a y floating around loose in the middle there? I could solve the first equation
for either variable, but I'd get fractions, and solving the second equation
for x would
also give me fractions. It wouldn't be "wrong" to make a different
choice, but it would probably be more difficult. Being lazy, I'll solve
the second equation for y:

4x + y = 24y = –4x + 24

Now I'll plug this in ("substitute
it") for "y"
in the first equation, and solve for x:

Now I can plug this x-value
back into either equation, and solve for y.
But since I already have an expression for "y =", it will be simplest to just
plug into this:

y = –4(5) + 24 = –20
+ 24 = 4

Then the solution
is (x, y) = (5, 4).

Warning: If I had substituted my "–4x + 24" expression into the same
equation as I'd used to solve for "y =", I would have gotten a true,
but useless, statement:

4x + (–4x + 24)
= 24
4x – 4x + 24 = 24
24 = 24

Twenty-four does equal twenty-four, but
who cares? So when using substitution, make sure you substitute into the other equation, or you'll just be wasting your time.

Solve the following
system by substitution.

y = 36 – 9x
3x + y/3 = 12

We already know (from the previous
lesson) that
these equations are actually both the same line; that is, this is a
dependent system. We know what this looks like graphically: we get two
identical line equations, and a graph with just one line displayed.
But what does this look like algebraically?

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The first equation is
already solved for y,
so I'll substitute that into the second equation:

3x + (36 – 9x)/3
= 12
3x + 12 – 3x = 12
12 = 12

Well, um... yes, twelve does equal
twelve, but so what?

I did substitute the first equation
into the second equation, so this unhelpful result is not because
of some screw-up on my part. It's just that this is what a dependent
system looks like when you try to find a solution. Remember that, when
you're trying to solve a system, you're trying to use the second equation
to narrow down the choices of points on the first equation. You're trying
to find the one single point that works in both equations. But in a
dependent system, the "second" equation is really just another
copy of the first equation, and all the points on the one line
will work in the other line.

In other words, I got an unhelpful result
because the second line equation didn't tell me anything new. This tells
me that the system is actually dependent, and that the solution is the
whole line:

solution: y = 36 – 9x

This is always true, by the way. When you
try to solve a system and you get a statement like "12
= 12" or "0
= 0" — something that's true,
but unhelpful (I mean, duh!, of course twelve equals twelve!) —
then you have a dependent system. We already knew, from the previous lesson,
that this system was dependent, but now you know what the algebra looks
like.

(Keep in mind that your text may format
the answer to look something like "(t,
36 – 9t)", or something
similar, using some variable, some "parameter", other than "x".
But this "parametrized" form of the solution means the exact
same thing as "the solution is the line y = 36 – 9x".)

Solve the following
system by substitution.

7x + 2y = 16
–21x – 6y = 24

Neither of these equations is particularly
easier than the other for solving. I'll get fractions, no matter
which equation and which variable I choose. So, um... I guess I'll take
the first equation, and I'll solve it for, um, y,
because at least the 2 (from the "2y")
will divide evenly into the 16.

7x + 2y = 16
2y = –7x + 16y = –( 7/2 )x + 8

Now I'll plug this into the other equation:

–21x – 6(–( 7/2 )x + 8) = 24
–21x + 21x – 48 = 24
–48
= 24

Um... I don't think so....

In this case, I got a nonsense result.
All my math was right, but I got an obviously wrong answer. So what
happened?

Keep in mind that, when solving, you're
trying to find where the lines intersect. What if they don't intersect?
Then you're going to get some kind of wrong answer when you assume that
there is a solution (as I did when I tried to find that solution). We
knew, from the previous lesson, that this system represents two parallel
lines. But I tried, by substitution, to find the intersection point
anyway. And I got a "garbage" result. Since there wasn't any
intersection point, my attempt led to utter nonsense.

solution: no solution
(inconsistent system)

This is always true, by
the way. When you get a nonsense result, this is the algebraic indication
that the system of equations is inconsistent.

Note that this is quite
different from the previous example. Warning: A true-but-useless result
(like "12 = 12")
is quite different from a nonsense "garbage" result (like "–48
= 24"), just as
two identical lines are quite different from two parallel lines. Don't
confuse the two. A useless result means a dependent system which has a
solution (the whole line); a nonsense result means an inconsistent system
which has no solution of any kind.